Proposition 11

Let G be a locally compact group, $\mu$ a left Haar measure on G. For a measure $\nu \neq 0$ on G to be left quasi-invariant, it is necessary and sufficient that $\nu$ be equivalent to $\mu$.

Sufficiency is obvious. Let $\nu \neq 0$ be a left quasi-invariant measure, and let us show that $\nu$ is equivalent to $\mu$. We can restrict ourselves to the case that $\nu > 0$. Let A be a compact subset of G. We will show, as will establish the proposition, that the conditions $\mu(A) = 0$, $\nu(A) = 0$ are equivalent (Ch. V, §5, No. 5, Th. 2).

a) For every $f \in \mathscr{K}_+(G)$, the function $(x, y) \mapsto f(x)\varphi_A(xy)$ on $G \times G$ is $(\nu \otimes \mu)$-integrable, because it is upper semi-continuous, bounded, and its support is contained in the compact set $K \times K^{-1}A$ if one sets $K = \operatorname{Supp} f$. Therefore, by the Lebesgue–Fubini theorem,

b) Suppose $\nu(A) = 0$. By hypothesis, $\nu(xA) = 0$ for all $x \in G$, therefore the right side of (36) is zero. Therefore there exists a $\nu$-negligible set $N_f$ such that, for $y \notin N_f$,

Let B be a compact subset of G such that $\nu(B) \neq 0$, and take for $f$ a function in $\mathscr{K}_+(G)$ equal to 1 on $AB^{-1}$. There then exists a $y \in B$ such that (37) is verified. But since $\varphi_A(x)f(xy^{-1}) = \varphi_A(x)$ for $y \in B$, this proves that $\mu(A) = 0$.

c) Suppose $\mu(A) = 0$. Then, for every $f \in \mathscr{K}_+(G)$, the left side of (36) is zero, hence also the right side. Consequently, there exists a locally $\mu$-negligible set M such that $\int \varphi_A(xy)\,d\nu(y) = 0$ for $x \notin M$. Since $\mu \neq 0$, it follows that $\nu(xA) = 0$ for some $x \in G$, whence $\nu(A) = 0$.

Applying Prop. 11 to $G^0$, one sees that the right quasi-invariant measures are identical with the left quasi-invariant measures. They are called simply quasi-invariant measures on G.